82.7k views
0 votes
A room of volume v contains air having equivalent molar mass m. if the temperature of the room is raised from t1 to t2, what mass of air will leave the room

User PDG
by
7.9k points

1 Answer

5 votes

The ideal gas law is:

pV = nRT

where p=pressure, V=volume, n=number of moles, R=gas constant, T=temperature, m=molar mass, M=mass

Setting up for M1 and M2 using the fact that n = M/m


pV1 = M1/m x RT1 ==> M1 = PmV1/RT1
pV2 = M2/m x RT2 ==> M2 = PmV2/RT2

Since volume doesn't change, V1 = V2 =V

The question is to find for the mass of air that enters the room, which is M2 – M1

M2 – M1 = (PmV2/RT2) – (PmV1/RT1 )

User Daniel Gasser
by
7.7k points